Difference between revisions of "2022 AMC 10A Problems/Problem 11"
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We are given that <cmath>2^m\cdot\sqrt{\frac{1}{4096}} = 2\cdot\sqrt[m]{\frac{1}{4096}}.</cmath> | We are given that <cmath>2^m\cdot\sqrt{\frac{1}{4096}} = 2\cdot\sqrt[m]{\frac{1}{4096}}.</cmath> | ||
− | Converting everything into powers of <math>2 | + | Converting everything into powers of <math>2</math> and equating exponents, we have |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
2^m\cdot(2^{-12})^{\frac12} &= 2\cdot (2^{-12})^{\frac1m} \ | 2^m\cdot(2^{-12})^{\frac12} &= 2\cdot (2^{-12})^{\frac1m} \ | ||
Line 14: | Line 14: | ||
m-6 &= 1-\frac{12}{m}. | m-6 &= 1-\frac{12}{m}. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | We multiply both sides by <math>m</math> | + | We multiply both sides by <math>m,</math> then rearrange as <cmath>m^2-7m+12=0.</cmath> |
By Vieta's Formulas, the sum of such values of <math>m</math> is <math>\boxed{\textbf{(C) } 7}.</math> | By Vieta's Formulas, the sum of such values of <math>m</math> is <math>\boxed{\textbf{(C) } 7}.</math> | ||
Note that <math>m=3</math> or <math>m=4</math> from the quadratic equation above. | Note that <math>m=3</math> or <math>m=4</math> from the quadratic equation above. | ||
− | ~MRENTHUSIASM | + | ~MRENTHUSIASM ~KingRavi |
− | |||
− | ~KingRavi | ||
==Solution 2 (Logarithms)== | ==Solution 2 (Logarithms)== | ||
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==Solution 3== | ==Solution 3== | ||
− | Since | + | Since third roots are conventionally positive integers, assume <math>m</math> is an integer, so <math>m</math> can only be <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>6</math>, and <math>12</math>. <math>\sqrt{\frac{1}{4096}}=\frac{1}{64}</math>. Testing out <math>m</math>, we see that only <math>3</math> and <math>4</math> work. Hence, <math>3+4=\boxed{\textbf{(C) }7}</math>. |
~MrThinker | ~MrThinker |
Latest revision as of 22:35, 10 November 2024
Contents
[hide]Problem
Ted mistakenly wrote as What is the sum of all real numbers for which these two expressions have the same value?
Solution 1
We are given that Converting everything into powers of and equating exponents, we have We multiply both sides by then rearrange as By Vieta's Formulas, the sum of such values of is
Note that or from the quadratic equation above.
~MRENTHUSIASM ~KingRavi
Solution 2 (Logarithms)
We can rewrite the equation using fractional exponents and take logarithms of both sides: We can then use the additive properties of logarithms to split them up: Using the power rule, the fact that and bringing the exponents down, we get from which or . Therefore, the answer is
- youtube.com/indianmathguy
Solution 3
Since third roots are conventionally positive integers, assume is an integer, so can only be , , , , , and . . Testing out , we see that only and work. Hence, .
~MrThinker
Video Solution 1
~Education, the Study of Everything
Video Solution 2
Video Solution 3
~Whiz
Video Solution by TheBeautyofMath
~IceMatrix
See Also
2022 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.